Multiplying and dividing by $\sqrt{x}$ yields
$$\frac{(x+7)^2\sqrt{x^2+2x}}{7x^3-2x^2}$$
where I would like to approximate the squareroot for sufficiently large $x$ with
$$\frac{(x+7)^2\sqrt{x^2+2x+1}}{7x^3-2x^2}=\frac{(x+7)^2(x+1)}{7x^3-2x^2}=\frac{x^3(49/x^3+63/x^2+15/x+1)}{x^3(7-2/x)}\longrightarrow 1/7.$$

Can anyone confirm that my approximation is valid and does anyone know how to solve this in a more "usual" way like I did?

It might help to notice that $x^2+2x+1=(x+1)^2$.
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icurays1Nov 12 '12 at 18:59

@icurays1: That was just a typo, thanks for reminding me of that!
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Christian IvicevicNov 12 '12 at 19:01

I think something is also funny with the way you've expanded the top. I think it should look like $x^3(49/x^3+63/x^2+15/x+1)$. Your limit is correct, though, you just need to be careful with the algebra.
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icurays1Nov 12 '12 at 19:05

2 Answers
2

Hint: A standard thing to do is to divide top and bottom by $x^2\sqrt{x}$.

The new top is
$$\left(1+\frac{7}{x}\right)^2 \sqrt{1+\frac{2}{x}},$$
and its behaviour for large $x$ is clear.

The new bottom is $7$ plus something tiny.

Remark: In the solution given in the OP, $\sqrt{x^2+2x}$ was replaced by $x+1$. True, this is fine, the change is indeed small. But if we are doing things formally, the replacement leaves a gap in the argument.

I did replace it with $\sqrt{x^2+2x+1}=\sqrt{(x+1)^2}=(x+1)$ thinking of sufficiently large $x$.
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Christian IvicevicNov 12 '12 at 19:32

@ChristianIvicevic: If we are informally trying to find the limit, what you did is just fine. We do need to develop some intuition about what changes will make no difference. However, if this is a formal exercise in finding a limit, any replacement needs to be fully justified, by showing it makes no difference.
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André NicolasNov 12 '12 at 19:51